Analysis of AlgorithmsAnalysis of Algorithmseinspem.upm.edu.my/wcaa/images/Algo_Analysis.pdf ·...

Analysis of AlgorithmsAnalysis of Algorithms

1. Asymptotic Notations2 Analysis of simple algorithms2. Analysis of simple algorithms

Analysis of Algorithms / Slide 2

Learning outcomesLearning outcomesYou should be able to:You should be able to:

Describe asymptotic notations: O Ω and ΘDescribe asymptotic notations: O, Ω , and Θ

Analyze the time complexity of algorithms


IntroductionIntroductionWhat is Algorithm?What is Algorithm?

a clearly specified set of simple instructions to be followed to solve a problem

Takes a set of values as input andTakes a set of values, as input and produces a value, or set of values, as output

May be specified I E li hIn EnglishAs a computer programAs a pseudo-code

Data structuresMethods of organizing data

Program = algorithms + data structuresProgram = algorithms + data structures


IntroductionIntroductionWhy need algorithm analysis ?Why need algorithm analysis ?

writing a working program is not good enoughThe program may be inefficient!The program may be inefficient!If the program is run on a large data set, then the running time becomes an issue


Example: Selection ProblemExample: Selection ProblemGiven a list of N numbers determine the kthGiven a list of N numbers, determine the kth largest, where k ≤ N.Algorithm 1:Algorithm 1:(1) Read N numbers into an array(2) Sort the array in decreasing order by some ( ) y g y

simple algorithm(3) Return the element in position k


Example: Selection ProblemExample: Selection Problem…Algorithm 2:Algorithm 2:(1) Read the first k elements into an array and sort

them in decreasing orderg(2) Each remaining element is read one by one

If smaller than the kth element, then it is ignoredOth i it i l d i it t t i thOtherwise, it is placed in its correct spot in the array, bumping one element out of the array.

(3) The element in the kth position is returned as the answer.


Example: Selection ProblemExample: Selection Problem…Which algorithm is better whenWhich algorithm is better when

N =100 and k = 100?N =100 and k = 1?N 100 and k 1?

What happens when N = 1,000,000 and k = 500,000?,There exist better algorithms


Algorithm AnalysisAlgorithm AnalysisWe only analyze correct algorithmsWe only analyze correct algorithmsAn algorithm is correct

If, for every input instance, it halts with the correct output

Incorrect algorithmsMight not halt at all on some input instancesMight halt with other than the desired answerMight halt with other than the desired answer

Analyzing an algorithmPredicting the resources that the algorithm requiresResources include

MemoryCommunication bandwidthComputational time (usually most important)


Algorithm AnalysisAlgorithm Analysis…Factors affecting the running timeFactors affecting the running time

computer compileralgorithm usedalgorithm usedinput to the algorithm

The content of the input affects the running timetypically the input size (number of items in the input) is the maintypically, the input size (number of items in the input) is the main consideration

E.g. sorting problem ⇒ the number of items to be sortedE.g. multiply two matrices together ⇒ the total number of

l t i th t t ielements in the two matrices

Machine model assumedInstructions are executed one after another, with no

t ti N t ll l tconcurrent operations ⇒ Not parallel computers


ExampleExampleCalculate ∑

N

i3Calculate ∑=i

i1

1 1

2

3

4

2N+2

4N

1

Lines 1 and 4 count for one unit eachLine 3: executed N times, each time four unitsLine 2: (1 for initialization, N+1 for all the tests, N for all the increments) total 2N + 2all the increments) total 2N + 2total cost: 6N + 4 ⇒ O(N)


Worst / average / best caseWorst- / average- / best-caseWorst-case running time of an algorithmWorst case running time of an algorithm

The longest running time for any input of size nAn upper bound on the running time for any input⇒ guarantee that the algorithm will never take longer⇒ guarantee that the algorithm will never take longerExample: Sort a set of numbers in increasing order; and the data is in decreasing orderThe worst case can occur fairly oftenThe worst case can occur fairly often

E.g. in searching a database for a particular piece of information

Best-case running timesort a set of numbers in increasing order; and the data issort a set of numbers in increasing order; and the data is already in increasing order

Average-case running timeMay be difficult to define what “average” meansMay be difficult to define what average means


Running time of algorithmsRunning-time of algorithmsBounds are for the algorithms rather thanBounds are for the algorithms, rather than programs

programs are just implementations of an algorithm,programs are just implementations of an algorithm, and almost always the details of the program do not affect the bounds

Bounds are for algorithms, rather thanblproblems

A problem can be solved with several algorithms, some are more efficient than otherssome are more efficient than others


Growth RateGrowth Rate

The idea is to establish a relative order among functions for large nfor large n∃ c , n0 > 0 such that f(N) ≤ c g(N) when N ≥ n0f(N) grows no faster than g(N) for “large” N


Asymptotic notation: Big OhAsymptotic notation: Big-Ohf(N) = O(g(N))f(N) = O(g(N))There are positive constants c and n0 such thatthat

f(N) ≤ c g(N) when N ≥ n0

The growth rate of f(N) is less than or equal tothe growth rate of g(N)g(N) is an upper bound on f(N)


Big Oh: exampleBig-Oh: exampleLet f(N) = 2N2 ThenLet f(N) = 2N2. Then

f(N) = O(N4)f(N) = O(N3)f(N) O(N )f(N) = O(N2) (best answer, asymptotically tight)

O(N2): reads “order N-squared” or “Big-Oh N-squared”


Big Oh: more examplesBig Oh: more examplesN2 / 2 – 3N = O(N2)N / 2 3N O(N )1 + 4N = O(N)7N2 + 10N + 3 = O(N2) = O(N3)log N log N / log 10 O(log N) O(log N)log10 N = log2 N / log2 10 = O(log2 N) = O(log N)sin N = O(1); 10 = O(1), 1010 = O(1)

)( 2NONNiN=⋅≤∑

)( 321

2 NONNiN

i=⋅≤∑ =

)(1

NONNii

=⋅≤∑ =

log N + N = O(N)logk N = O(N) for any constant kN = O(2N) but 2N is not O(N)N = O(2N), but 2N is not O(N)210N is not O(2N)


Math Re ie : logarithmic f nctionsMath Review: logarithmic functionsabiffbxa log ==

bbaab

abiffbx x

loglogloglog

log+=

abb

bm

ma log

loglog =

naaba

an

b loglogloglog =

=

xdaaa bbb

1loglog)(loglog ≠=

xdxxd e 1log=


Some rulesSome rulesWhen considering the growth rate of a function usingWhen considering the growth rate of a function using Big-OhIgnore the lower order terms and the coefficients of gthe highest-order termNo need to specify the base of logarithm

Ch i th b f t t t th h thChanging the base from one constant to another changes the value of the logarithm by only a constant factor

If T1(N) = O(f(N) and T2(N) = O(g(N)), thenT1(N) + T2(N) = max(O(f(N)), O(g(N))),T1(N) * T2(N) = O(f(N) * g(N))


Big OmegaBig-Omega

∃ c , n0 > 0 such that f(N) ≥ c g(N) when N ≥ n0

f(N) grows no slower than g(N) for “large” N


Big OmegaBig-Omega

f(N) = Ω(g(N))There are positive constants c and n suchThere are positive constants c and n0 such that

f(N) ≥ c g(N) when N ≥ nf(N) ≥ c g(N) when N ≥ n0

The growth rate of f(N) is greater than or g ( ) gequal to the growth rate of g(N).


Big Omega: examplesBig-Omega: examplesLet f(N) = 2N2 ThenLet f(N) = 2N2. Then

f(N) = Ω(N)f(N) = Ω(N2) (best answer)f(N) Ω(N ) (best answer)


f(N) = Θ(g(N))f(N) = Θ(g(N))

the growth rate of f(N) is the same as the growth rate f (N)of g(N)


Big ThetaBig-Theta

f(N) = Θ(g(N)) iff f(N) = O(g(N)) and f(N) = Ω(g(N))f(N) O(g(N)) and f(N) Ω(g(N))The growth rate of f(N) equals the growth rate of g(N)Example: Let f(N)=N2 , g(N)=2N2

Since f(N) = O(g(N)) and f(N) = Ω(g(N)), th f(N) ( (N))thus f(N) = Θ(g(N)).

Big-Theta means the bound is the tightest possiblepossible.


Some rulesSome rules

If T(N) is a polynomial of degree k, then T(N) = Θ(Nk)T(N) = Θ(N ).

For logarithmic functionsFor logarithmic functions,T(logm N) = Θ(log N).


Typical Growth Rates


Growth ratesGrowth rates …Doubling the input sizeDoubling the input size

f(N) = c ⇒ f(2N) = f(N) = cf(N) = log N ⇒ f(2N) = f(N) + log 2f(N) = N ⇒ f(2N) = 2 f(N) f(N) = N2 ⇒ f(2N) = 4 f(N) f(N) = N3 ⇒ f(2N) = 8 f(N)f(N) = N3 ⇒ f(2N) = 8 f(N) f(N) = 2N ⇒ f(2N) = f2(N)

Advantages of algorithm analysisAdvantages of algorithm analysis To eliminate bad algorithms earlypinpoints the bottlenecks, which are worth coding

f llcarefully


Using L' Hopital's ruleUsing L Hopital s ruleL' Hopital's ruleL Hopital s rule

If and

then)(lim Nf )(lim Nf ′

∞=∞→

)(lim Nfn

∞=∞→

)(lim Ngn

then =

Determine the relative growth rates (using L' Hopital's rule if

)(lim

Ngn ∞→ )(lim

Ngn ′∞→

necessary)compute

)()(lim

NgNf

n ∞→

if 0: f(N) = O(g(N)) and f(N) is not Θ(g(N))if constant ≠ 0: f(N) = Θ(g(N))if ∞: f(N) = Ω(f(N)) and f(N) is not Θ(g(N))if ∞: f(N) = Ω(f(N)) and f(N) is not Θ(g(N))limit oscillates: no relation


General RulesGeneral RulesFor loops

at most the running time of the statements inside the for-loop (including tests) times the number of iterations.iterations.

Nested for loops

the running time of the statement multiplied by the product of the sizes of all the for-loops.O(N2)


General rules (cont’d)General rules (cont d)Consecutive statementsConsecutive statements

These just addO(N) + O(N2) = O(N2)O(N) + O(N ) O(N )

If S1Else S2

never more than the running time of the test plus the larger of the running times of S1 and S2.


Another ExampleAnother ExampleMaximum Subsequence Sum ProblemMaximum Subsequence Sum ProblemGiven (possibly negative) integers A1, A2, ...., A find the maximum value of ∑

j

AAn, find the maximum value of

For convenience the maximum subsequence sum

∑=ik

kA

For convenience, the maximum subsequence sum is 0 if all the integers are negative

E.g. for input –2, 11, -4, 13, -5, -2Answer: 20 (A2 through A4)


Algorithm 1: SimpleAlgorithm 1: SimpleExhaustively tries all possibilities (brute force)Exhaustively tries all possibilities (brute force)

O(N3)


Algorithm 2: Divide and conquerAlgorithm 2: Divide-and-conquer Divide-and-conquerDivide and conquer

split the problem into two roughly equal subproblems, which are then solved recursivelypatch together the two solutions of the subproblems to arrivepatch together the two solutions of the subproblems to arrive at a solution for the whole problem

The maximum subsequence sum can be Entirely in the left half of the inputEntirely in the right half of the inputIt crosses the middle and is in both halves


Algorithm 2 (cont’d)Algorithm 2 (cont d)

The first two cases can be solved recursivelyFor the last case:For the last case:

find the largest sum in the first half that includes the last element in the first halfthe largest sum in the second half that includes the firstthe largest sum in the second half that includes the first element in the second halfadd these two sums together


Algorithm 2Algorithm 2 …

O(1)

T(m/2)

T(m/2)

O(m)

T(m/2)

O(1)


Algorithm 2 (cont’d)Algorithm 2 (cont d)Recurrence equationRecurrence equation

T =1)1(

NNTNT += )2

(2)(

2 T(N/2): two subproblems, each of size N/2N: for “patching” two solutions to find solution toN: for patching two solutions to find solution to whole problem


Algorithm 2 (cont’d)Algorithm 2 (cont d)Solving the recurrence: NNTNT += )

2(2)(Solving the recurrence:

NNT += 2)4

(4

)2

()(

NNT

=

+= 3)8

(8

L

With k=log N (i.e. 2k = N), we havekNNT k

k += )2

(2

Thus the running time is O(N log N)NNN

NNTNNT+=+=

loglog)1()(

Thus, the running time is O(N log N)faster than Algorithm 1 for large data sets


QuestionQuestion

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